$$\eqalign{ & {\text{Clearly,}}\,\,\frac{b}{{{x_1} - a}} = \tan \,\theta = \frac{{{y_1} - b}}{a} \cr & \therefore \,\,\,ab = {x_1}{y_1} - b{x_1} - a{y_1} + ab \cr & \Rightarrow \frac{a}{{{x_1}}} + \frac{b}{{{y_1}}} = 1 \cr} $$

The pair $${y^2} - 4xy - {x^2} = 0$$     contains lines which are at right angles because $$a+b=0.$$

It is given that $$\alpha ,\,\beta $$  are the roots of the equation $${x^2} - 6{p_1}x + 2 = 0$$
$$\therefore \,\alpha + \beta = 6{p_1},\,\alpha \beta = 2......\left( {\text{i}} \right)$$
$$\beta ,\,\gamma $$  are the roots of the equation $${x^2} - 6{p_2}x + 3 = 0$$
$$\therefore \,\beta + \gamma = 6{p_2},\,\beta \gamma = 3......\left( {{\text{ii}}} \right)$$
$$\gamma ,\,\alpha $$  are the roots of the equation $${x^2} - 6{p_3}x + 6 = 0$$
$$\therefore \,\gamma + \alpha = 6{p_3},\,\gamma \alpha = 6......\left( {{\text{iii}}} \right)$$
From Equations $$\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right)$$   and $$\left( {{\text{iii}}} \right)$$  we get
$$\eqalign{ & \Rightarrow \alpha \beta \gamma = 6\,\,\,\,\,\left[ {\therefore \,\alpha ,\,\beta ,\,\gamma > 0} \right] \cr & {\text{Now, }}\alpha \beta = 2{\text{ and }}\alpha \beta \gamma = 6 \cr & \Rightarrow \gamma = 3 \cr & \beta \gamma = 3{\text{ }}\,\,{\text{and}}\,\,{\text{ }}\alpha \beta \gamma = 6 \cr & \alpha = 3,\,\,{\text{and }}\alpha = 6\alpha \beta \gamma = 6 \cr & \Rightarrow \beta = 1 \cr & \therefore \alpha + \beta = 6{p_1} \cr & \Rightarrow 3 = 6{p_1} \cr & \Rightarrow {p_1} = \frac{1}{2} \cr & \beta + \gamma = 6{p_2} \cr & \Rightarrow 4 = 6{p_2} \cr & \Rightarrow {p_2} = \frac{2}{3} \cr & {\text{and }}\,\gamma + \alpha = 6{p_3} \cr & \Rightarrow 5 = 6{p_3} \cr & \Rightarrow {p_3} = \frac{5}{6} \cr} $$
The coordinates of the centroid of triangle are
$$\eqalign{ & \left( {\frac{{\alpha + \beta + \gamma }}{3},\,\frac{1}{3}\left( {\frac{1}{\alpha } + \frac{1}{\beta } + \frac{1}{\gamma }} \right)} \right) \cr & {\text{or }}\left( {\frac{6}{3},\,\frac{1}{3}\left( {\frac{1}{2} + 1 + \frac{1}{3}} \right)} \right) \cr & {\text{or }}\left( {2,\,\frac{{11}}{{18}}} \right) \cr} $$

Here $$y\left( {{y^2} - 4{x^2}} \right) = 0$$    gives the lines $$y=0,\, y=2x$$    and $$y=-2x,$$   which are concurrent at (0, 0).

$$\eqalign{ & y = \left( {2 + \sqrt 3 } \right)x + 4......\left( 1 \right) \cr & y = k\alpha + 6......\left( 2 \right) \cr & {\text{By equation }}\left( 1 \right) \Rightarrow {m_1} = \left( {2 + \sqrt 3 } \right)\,; \cr & {\text{By equation }}\left( 2 \right) \Rightarrow {m_2} = k \cr & \therefore \,\tan \,{60^ \circ } = \frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} \Rightarrow k = - 1 \cr} $$

$$\eqalign{ & {\text{Let }}P{\text{ be on }}x + 2y = 1 \cr & \Rightarrow 1 + \frac{t}{{\sqrt 2 }} + 2\left( {2 + \frac{t}{{\sqrt 2 }}} \right) = 1{\text{ or }}t = - \frac{{4\sqrt 2 }}{3} \cr & {\text{Let }}P{\text{ be on 2}}x + 4y = 15 \cr & \Rightarrow 2\left( {1 + \frac{t}{{\sqrt 2 }}} \right) + 4\left( {2 + \frac{t}{{\sqrt 2 }}} \right) = 15{\text{ or }}t = \frac{{5\sqrt 2 }}{6} \cr} $$

The perpendicular distance between the parallel lines in pairs is same, hence the lines form a rhombus.

We have, $$2{x^2} + 5xy + 3{y^2} + 6x + 7y + 4 = 0$$
Comparing this equation with $$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0,$$         we get $$a = 2,\,b = 3,\,h = \frac{5}{2}$$
$$\eqalign{ & \therefore \,\tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}} \cr & = \frac{{2\sqrt {\frac{{25}}{4} - 2 \times 3} }}{{2 + 3}} \cr & = \frac{{2\sqrt {\frac{1}{4}} }}{5} \cr & = \frac{{2 \times \frac{1}{2}}}{5} \cr & = \frac{1}{5}\tan \,\theta = \frac{1}{5} \cr & \Rightarrow m = \frac{1}{5} \cr} $$

$$\eqalign{ & {\text{We have}} \cr & P = \left( {1,\,0} \right),\,Q = \left( { - 1,\,0} \right),\,R = \left( {2,\,0} \right) \cr & {\text{Let}}\,\,\,{\text{ }}S = \left( {x,\,y} \right) \cr & {\text{ATQ}}\,\,\,S{Q^2} + S{R^2} = 2S{P^2} \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} + {\left( {x - 2} \right)^2} + {y^2} = 2\left[ {{{\left( {x - 1} \right)}^2} + {y^2}} \right] \cr & \Rightarrow 2{x^2} + 2{y^2} - 2x + 5 = 2{x^2} + 2{y^2} - 4x + 2 \cr & \Rightarrow 2x + 3 = 0 \cr & \Rightarrow x = - \frac{3}{2} \cr} $$
Which is a straight line parallel to $$y$$-axis.

Two joining points are $$\left( {p,\,q} \right)$$  and $$\left( {q,\, - p} \right)$$
Mid point of $$\left( {p,\,q} \right)$$  and $$\left( {q,\, - p} \right)$$  is $$\left( {\frac{{p + q}}{2},\,\frac{{q - p}}{2}} \right)$$
But it is given that the mid-point is $$\left( {\frac{r}{2},\,\frac{s}{2}} \right).$$
$$\eqalign{ & \therefore \,\frac{{p + q}}{2} = \frac{r}{2}{\text{ and }}\frac{{q - p}}{2} = \frac{s}{2} \cr & \Rightarrow p + q = r{\text{ and }}q - p = s \cr} $$
Now, length of segment
$$\eqalign{ & = \sqrt {{{\left( {p - q} \right)}^2} + {{\left( {q + p} \right)}^2}} \,\,\,\,\,\left( {{\text{by distance formula}}} \right) \cr & = \sqrt {{s^2} + {r^2}} \cr} $$